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De Moivre's Theorem - IB Mathematics Analysis and Approaches

  1. Let z=cos⁡θ+isin⁡θz=\cos\theta+i\sin\theta. Find the imaginary part of 11−z\frac{1}{1-z}.
    Answer: 12cot⁡(θ2)\frac{1}{2}\cot(\frac{\theta}{2}) We have z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta. So, 1−z=1−(cos⁡θ+isin⁡θ)=(1−cos⁡θ)−isin⁡θ1-z = 1 - (\cos\theta + i\sin\theta) = (1-\cos\theta) - i\sin\theta. To find 11−z\frac{1}{1-z}, we multiply the numerator and denominator by the conjugate of the denominator, which is (1−cos⁡θ)+isin⁡θ(1-\cos\theta) + i\sin\theta. 1(1−cos⁡θ)−isin⁡θ×(1−cos⁡θ)+isin⁡θ(1−cos⁡θ)+isin⁡θ=(1−cos⁡θ)+isin⁡θ(1−cos⁡θ)2+sin⁡2θ\frac{1}{(1-\cos\theta) - i\sin\theta} \times \frac{(1-\cos\theta) + i\sin\theta}{(1-\cos\theta) + i\sin\theta} = \frac{(1-\cos\theta) + i\sin\theta}{(1-\cos\theta)^2 + \sin^2\theta}. The denominator is 1−2cos⁡θ+cos⁡2θ+sin⁡2θ=1−2cos⁡θ+1=2−2cos⁡θ=2(1−cos⁡θ)1 - 2\cos\theta + \cos^2\theta + \sin^2\theta = 1 - 2\cos\theta + 1 = 2 - 2\cos\theta = 2(1-\cos\theta). So the expression is 1−cos⁡θ2(1−cos⁡θ)+isin⁡θ2(1−cos⁡θ)=12+isin⁡θ2(1−cos⁡θ)\frac{1-\cos\theta}{2(1-\cos\theta)} + i\frac{\sin\theta}{2(1-\cos\theta)} = \frac{1}{2} + i\frac{\sin\theta}{2(1-\cos\theta)}. The imaginary part is sin⁡θ2(1−cos⁡θ)\frac{\sin\theta}{2(1-\cos\theta)}. Using half-angle identities, sin⁡θ=2sin⁡(θ2)cos⁡(θ2)\sin\theta = 2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2}) and 1−cos⁡θ=2sin⁡2(θ2)1-\cos\theta = 2\sin^2(\frac{\theta}{2}). So the imaginary part is 2sin⁡(θ2)cos⁡(θ2)2(2sin⁡2(θ2))=cos⁡(θ2)2sin⁡(θ2)=12cot⁡(θ2)\frac{2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}{2(2\sin^2(\frac{\theta}{2}))} = \frac{\cos(\frac{\theta}{2})}{2\sin(\frac{\theta}{2})} = \frac{1}{2}\cot(\frac{\theta}{2}).
  2. Let z=eiθz = e^{i\theta}. Find the modulus of the complex number w=1+z+z2w = 1+z+z^2.
    Answer: ∣sin⁡(3θ/2)/sin⁡(θ/2)∣|\sin(3\theta/2)/\sin(\theta/2)| The expression w=1+z+z2w = 1+z+z^2 is a geometric series with first term a=1a=1, common ratio zz, and n=3n=3 terms. The sum is w=z3−1z−1w = \frac{z^3-1}{z-1}. Substitute z=eiθz = e^{i\theta}: w=ei3θ−1eiθ−1w = \frac{e^{i3\theta}-1}{e^{i\theta}-1}. We can use the identity eiα−1=eiα/2(eiα/2−e−iα/2)=eiα/2(2isin⁡(α/2))e^{i\alpha}-1 = e^{i\alpha/2}(e^{i\alpha/2}-e^{-i\alpha/2}) = e^{i\alpha/2}(2i\sin(\alpha/2)). Applying this to the numerator and denominator: Numerator: ei3θ−1=ei3θ/2(2isin⁡(3θ/2))e^{i3\theta}-1 = e^{i3\theta/2}(2i\sin(3\theta/2)). Denominator: eiθ−1=eiθ/2(2isin⁡(θ/2))e^{i\theta}-1 = e^{i\theta/2}(2i\sin(\theta/2)). So, w=ei3θ/2(2isin⁡(3θ/2))eiθ/2(2isin⁡(θ/2))=ei(3θ/2−θ/2)sin⁡(3θ/2)sin⁡(θ/2)=eiθsin⁡(3θ/2)sin⁡(θ/2)w = \frac{e^{i3\theta/2}(2i\sin(3\theta/2))}{e^{i\theta/2}(2i\sin(\theta/2))} = e^{i(3\theta/2 - \theta/2)} \frac{\sin(3\theta/2)}{\sin(\theta/2)} = e^{i\theta} \frac{\sin(3\theta/2)}{\sin(\theta/2)}. We want to find the modulus ∣w∣|w|. Since ∣eiθ∣=1|e^{i\theta}|=1, we have: ∣w∣=∣eiθ∣∣sin⁡(3θ/2)sin⁡(θ/2)∣=1⋅∣sin⁡(3θ/2)sin⁡(θ/2)∣=∣sin⁡(3θ/2)sin⁡(θ/2)∣|w| = |e^{i\theta}| \left| \frac{\sin(3\theta/2)}{\sin(\theta/2)} \right| = 1 \cdot \left| \frac{\sin(3\theta/2)}{\sin(\theta/2)} \right| = \left| \frac{\sin(3\theta/2)}{\sin(\theta/2)} \right|.
  3. Let ω=ei2π/9\omega = e^{i2\pi/9}. Find the value of the sum ∑k=08(ωk)2\sum_{k=0}^{8} (\omega^k)^2.
    Answer: 0 The sum can be rewritten as ∑k=08(ω2)k\sum_{k=0}^{8} (\omega^2)^k. This is a finite geometric series with first term a=(ω2)0=1a = (\omega^2)^0 = 1, common ratio r=ω2r = \omega^2, and n=9n=9 terms (from k=0 to k=8). The sum of a finite geometric series is given by Sn=arn−1r−1S_n = a\frac{r^n-1}{r-1}. Here, S9=1⋅(ω2)9−1ω2−1=ω18−1ω2−1S_9 = 1 \cdot \frac{(\omega^2)^9 - 1}{\omega^2 - 1} = \frac{\omega^{18} - 1}{\omega^2 - 1}. We are given that ω=ei2π/9\omega = e^{i2\pi/9}, which is a 9th root of unity. This means ω9=1\omega^9 = 1. Therefore, ω18=(ω9)2=12=1\omega^{18} = (\omega^9)^2 = 1^2 = 1. The numerator of the sum becomes 1−1=01 - 1 = 0. The denominator is ω2−1\omega^2 - 1. Since ω\omega is a primitive 9th root of unity, ω2≠1\omega^2 \neq 1, so the denominator is not zero. Thus, the value of the sum is 0ω2−1=0\frac{0}{\omega^2 - 1} = 0.
  4. If nn is a positive integer, what is the value of (1+i1−i)2n+(1−i1+i)2n (\frac{1+i}{1-i})^{2n} + (\frac{1-i}{1+i})^{2n} ?
    Answer: 2(−1)n2(-1)^n First, simplify the expressions inside the parentheses. 1+i1−i=1+i1−i×1+i1+i=(1+i)212−i2=1+2i+i21−(−1)=1+2i−12=2i2=i\frac{1+i}{1-i} = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2-i^2} = \frac{1+2i+i^2}{1-(-1)} = \frac{1+2i-1}{2} = \frac{2i}{2} = i. 1−i1+i=1i=−i\frac{1-i}{1+i} = \frac{1}{i} = -i. Now substitute these simplified forms back into the original expression: (i)2n+(−i)2n (i)^{2n} + (-i)^{2n} . This can be written as (i2)n+((−i)2)n(i^2)^n + ((-i)^2)^n. Since i2=−1i^2 = -1 and (−i)2=i2=−1(-i)^2 = i^2 = -1, the expression becomes: (−1)n+(−1)n=2(−1)n (-1)^n + (-1)^n = 2(-1)^n . This value is 2 if nn is even, and -2 if nn is odd.
  5. Find the root of z3=8iz^3 = 8i that has a negative real part.
    Answer: −3+i-\sqrt{3}+i When solving equations of the form zn=wz^n = w where ww is a complex number, you're finding the nn-th roots of ww. The key is to express ww in polar form, then use the formula for complex roots. First, convert 8i8i to polar form. Since 8i=0+8i8i = 0 + 8i, we have r=∣8i∣=8r = |8i| = 8 and θ=π2\theta = \frac{\pi}{2} (since it lies on the positive imaginary axis). So 8i=8eiπ/28i = 8e^{i\pi/2}. For the equation z3=8iz^3 = 8i, the three cube roots are: zk=83⋅ei(π/2+2πk)/3z_k = \sqrt[3]{8} \cdot e^{i(\pi/2 + 2\pi k)/3} where k=0,1,2k = 0, 1, 2 Since 83=2\sqrt[3]{8} = 2, we get: - z0=2eiπ/6=2(cos⁡(π/6)+isin⁡(π/6))=2(32+i2)=3+iz_0 = 2e^{i\pi/6} = 2(\cos(\pi/6) + i\sin(\pi/6)) = 2(\frac{\sqrt{3}}{2} + \frac{i}{2}) = \sqrt{3} + i - z1=2ei5π/6=2(−32+i2)=−3+iz_1 = 2e^{i5\pi/6} = 2(-\frac{\sqrt{3}}{2} + \frac{i}{2}) = -\sqrt{3} + i - z2=2ei3π/2=2(0−i)=−2iz_2 = 2e^{i3\pi/2} = 2(0 - i) = -2i The question asks for the root with negative real part. Among our roots, z1=−3+iz_1 = -\sqrt{3} + i has real part −3<0-\sqrt{3} < 0, making D correct. Option A (−2i-2i) has zero real part, not negative. Option B (3+i\sqrt{3} + i) has positive real part 3\sqrt{3}. Option C (3−i\sqrt{3} - i) also has positive real part and isn't even a root of our equation. Study tip: Always find all roots systematically using the polar form method, then select based on the given condition. Don't guess—calculate all possibilities first.
  6. Let ω\omega be a non-real cube root of unity. Find the value of (1+ω−ω2)3(1+\omega-\omega^2)^3.
    Answer: -8 The cube roots of unity satisfy the equation z3−1=0z^3-1=0, which means (z−1)(z2+z+1)=0(z-1)(z^2+z+1)=0. Since ω\omega is a non-real root, it must be a root of z2+z+1=0z^2+z+1=0. Therefore, we have the key properties: 1+ω+ω2=01+\omega+\omega^2=0 and ω3=1\omega^3=1. From 1+ω+ω2=01+\omega+\omega^2=0, we can write 1+ω=−ω21+\omega = -\omega^2. Substitute this into the expression: (1+ω−ω2)3=(−ω2−ω2)3=(−2ω2)3(1+\omega-\omega^2)^3 = (-\omega^2 - \omega^2)^3 = (-2\omega^2)^3. Now, we simplify: (−2ω2)3=(−2)3(ω2)3=−8ω6(-2\omega^2)^3 = (-2)^3 (\omega^2)^3 = -8 \omega^6. Since ω3=1\omega^3=1, we have ω6=(ω3)2=12=1\omega^6 = (\omega^3)^2 = 1^2 = 1. Therefore, the final value is −8×1=−8-8 \times 1 = -8.
  7. Let z1,z2,z3z_1, z_2, z_3 be the cube roots of 8i8i. Find the value of z12+z22+z32z_1^2 + z_2^2 + z_3^2.
    Answer: 0 When you encounter problems about roots of complex numbers, think about using polar form and the relationship between roots and coefficients of polynomials. This question tests both concepts. First, let's find the cube roots of 8i8i. Converting to polar form: 8i=8ei(π/2+2πk)8i = 8e^{i(\pi/2 + 2\pi k)} for integer kk. The cube roots are: zk=2ei(π/6+2πk/3)z_k = 2e^{i(\pi/6 + 2\pi k/3)} for k=0,1,2k = 0, 1, 2 This gives us: - z1=2eiπ/6=3+iz_1 = 2e^{i\pi/6} = \sqrt{3} + i - z2=2ei5π/6=−3+iz_2 = 2e^{i5\pi/6} = -\sqrt{3} + i - z3=2ei3π/2=−2iz_3 = 2e^{i3\pi/2} = -2i However, there's a more elegant approach using Vieta's formulas. Since z1,z2,z3z_1, z_2, z_3 are roots of w3=8iw^3 = 8i, they satisfy w3−8i=0w^3 - 8i = 0. By Vieta's formulas: z1+z2+z3=0z_1 + z_2 + z_3 = 0 and z1z2z3=8iz_1z_2z_3 = 8i. Using the identity (z1+z2+z3)2=z12+z22+z32+2(z1z2+z1z3+z2z3)(z_1 + z_2 + z_3)^2 = z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_1z_3 + z_2z_3): Since z1+z2+z3=0z_1 + z_2 + z_3 = 0, we get z12+z22+z32=−2(z1z2+z1z3+z2z3)=0z_1^2 + z_2^2 + z_3^2 = -2(z_1z_2 + z_1z_3 + z_2z_3) = 0. Answer (A) -4 likely comes from incorrectly computing z1z2+z1z3+z2z3z_1z_2 + z_1z_3 + z_2z_3. Answer (B) 4i might result from confusing this with z1z2z3z_1z_2z_3. Answer (C) 4 could come from taking the modulus incorrectly. Key strategy: For problems involving multiple roots of complex numbers, consider using Vieta's formulas alongside the symmetry properties—this often leads to elegant solutions without computing individual roots.
  8. Using De Moivre's theorem, the trigonometric identity for sin⁡(3θ)\sin(3\theta) can be expressed in terms of sin⁡θ\sin\theta. Which of the following is the correct expression?
    Answer: 3sin⁡θ−4sin⁡3θ3\sin\theta - 4\sin^3\theta By De Moivre's theorem, cos⁡(3θ)+isin⁡(3θ)=(cos⁡θ+isin⁡θ)3\cos(3\theta) + i\sin(3\theta) = (\cos\theta + i\sin\theta)^3. We expand the right side using the binomial theorem: (cos⁡θ)3+3(cos⁡θ)2(isin⁡θ)+3(cos⁡θ)(isin⁡θ)2+(isin⁡θ)3(\cos\theta)^3 + 3(\cos\theta)^2(i\sin\theta) + 3(\cos\theta)(i\sin\theta)^2 + (i\sin\theta)^3. This simplifies to: cos⁡3θ+3icos⁡2θsin⁡θ−3cos⁡θsin⁡2θ−isin⁡3θ\cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta. Group the real and imaginary parts: (cos⁡3θ−3cos⁡θsin⁡2θ)+i(3cos⁡2θsin⁡θ−sin⁡3θ)(\cos^3\theta - 3\cos\theta\sin^2\theta) + i(3\cos^2\theta\sin\theta - \sin^3\theta). Equating the imaginary parts gives: sin⁡(3θ)=3cos⁡2θsin⁡θ−sin⁡3θ\sin(3\theta) = 3\cos^2\theta\sin\theta - \sin^3\theta. This is option D, but it's not fully in terms of sin⁡θ\sin\theta. To express it purely in terms of sin⁡θ\sin\theta, we use the identity cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta:sin⁡(3θ)=3(1−sin⁡2θ)sin⁡θ−sin⁡3θ=3sin⁡θ−3sin⁡3θ−sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin(3\theta) = 3(1 - \sin^2\theta)\sin\theta - \sin^3\theta = 3\sin\theta - 3\sin^3\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta.
  9. Find the real part of (1+i3)−3(1+i\sqrt{3})^{-3}.
    Answer: -1/8 Let z=1+i3z = 1+i\sqrt{3}. First, convert zz to polar form. The modulus is r=∣z∣=12+(3)2=1+3=2r = |z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2. The argument θ\theta satisfies cos⁡θ=1/2\cos\theta=1/2 and sin⁡θ=3/2\sin\theta=\sqrt{3}/2, so θ=π/3\theta = \pi/3. Thus, z=2eiπ/3z = 2e^{i\pi/3}. We need to find z−3z^{-3}. Using De Moivre's theorem: z−3=(2eiπ/3)−3=2−3ei(−3)(π/3)=18e−iπz^{-3} = (2e^{i\pi/3})^{-3} = 2^{-3}e^{i(-3)(\pi/3)} = \frac{1}{8}e^{-i\pi}. Now, convert back to Cartesian form: 18e−iπ=18(cos⁡(−π)+isin⁡(−π))=18(−1+i(0))=−18\frac{1}{8}e^{-i\pi} = \frac{1}{8}(\cos(-\pi) + i\sin(-\pi)) = \frac{1}{8}(-1 + i(0)) = -\frac{1}{8}. The result is a real number, so its real part is −1/8-1/8.
  10. Find the product of the non-real roots of the equation z6−64=0z^6 - 64 = 0.
    Answer: 16 The equation is z6=64z^6 = 64. The roots are the 6th roots of 64. The modulus of each root is 646=2\sqrt[6]{64} = 2. The arguments are 2kπ6=kπ3\frac{2k\pi}{6} = \frac{k\pi}{3} for k=0,1,...,5k=0, 1, ..., 5. The roots are 2,2eiπ/3,2ei2π/3,2eiπ,2ei4π/3,2ei5π/32, 2e^{i\pi/3}, 2e^{i2\pi/3}, 2e^{i\pi}, 2e^{i4\pi/3}, 2e^{i5\pi/3}. In Cartesian form, the roots are 2,1+i3,−1+i3,−2,−1−i3,1−i32, 1+i\sqrt{3}, -1+i\sqrt{3}, -2, -1-i\sqrt{3}, 1-i\sqrt{3}. The real roots are 22 and −2-2. The four non-real roots are 1±i31\pm i\sqrt{3} and −1±i3-1\pm i\sqrt{3}. Let PP be the product of all roots. For a polynomial anzn+...+a0=0a_nz^n + ... + a_0 = 0, the product of roots is (−1)na0/an(-1)^n a_0/a_n. Here, P=(−1)6(−64)/1=−64P = (-1)^6 (-64)/1 = -64. The product of all roots is the product of the real roots multiplied by the product of the non-real roots. Product of real roots = 2×(−2)=−42 \times (-2) = -4. Product of non-real roots PNRP_{NR} is such that P=PR×PNRP = P_R \times P_{NR}. −64=−4×PNR-64 = -4 \times P_{NR}. PNR=−64−4=16P_{NR} = \frac{-64}{-4} = 16.
  11. Let z=−1−iz = -1-i. When plotted on an Argand diagram, the roots of w4=zw^4 = z form a square. Find the length of a diagonal of this square.
    Answer: 29/82^{9/8} First, represent z=−1−iz = -1-i in polar form. The modulus is ∣z∣=(−1)2+(−1)2=2|z| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}. The argument is θ=−3π/4\theta = -3\pi/4 (since it's in the third quadrant). So, z=2e−i3π/4z = \sqrt{2} e^{-i3\pi/4}. The roots of w4=zw^4 = z are given by wk=(∣z∣)1/4ei(θ+2kπ)/4w_k = (|z|)^{1/4} e^{i(\theta+2k\pi)/4} for k=0,1,2,3k=0,1,2,3. The modulus of each root is ∣wk∣=(2)1/4=(21/2)1/4=21/8|w_k| = (\sqrt{2})^{1/4} = (2^{1/2})^{1/4} = 2^{1/8}. Geometrically, the four roots lie on a circle centered at the origin with radius R=21/8R = 2^{1/8}. These roots form the vertices of a square inscribed in this circle. The diagonal of this square is a diameter of the circle. The length of the diameter is 2R2R. Length of diagonal = 2×21/8=21×21/8=21+1/8=29/82 \times 2^{1/8} = 2^1 \times 2^{1/8} = 2^{1+1/8} = 2^{9/8}.
  12. The complex number z=3+iz = \sqrt{3}+i is a root of the equation zn=kz^n=k, where kk is a real number and nn is a positive integer. Which of the following could be the value of kk?
    Answer: -64 First, convert z=3+iz = \sqrt{3}+i into polar form. The modulus is r=∣z∣=(3)2+12=3+1=2r = |z| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2. The argument θ\theta satisfies cos⁡θ=3/2\cos\theta = \sqrt{3}/2 and sin⁡θ=1/2\sin\theta = 1/2, so θ=π/6\theta = \pi/6. Thus, z=2eiπ/6z=2e^{i\pi/6}. We are given zn=kz^n=k. Using De Moivre's theorem: k=(2eiπ/6)n=2neinπ/6k = (2e^{i\pi/6})^n = 2^n e^{in\pi/6}. Since kk is a real number, the imaginary part of kk must be zero. This means the argument of kk must be an integer multiple of π\pi. So, nπ6=mπ\frac{n\pi}{6} = m\pi for some integer mm. This simplifies to n=6mn=6m. Since nn must be a positive integer, the smallest possible value for nn is 6 (when m=1m=1). Let's find kk for n=6n=6: k=26ei(6π/6)=64eiπ=64(−1)=−64k = 2^6 e^{i(6\pi/6)} = 64e^{i\pi} = 64(-1) = -64. If we take m=2m=2, then n=12n=12, and k=212ei(12π/6)=212ei2π=4096k=2^{12}e^{i(12\pi/6)} = 2^{12}e^{i2\pi} = 4096. This is not an option. Of the given choices, only -64 is a possible value for kk.
  13. Let zkz_k for k=1,2,3,4,5k=1, 2, 3, 4, 5 be the five distinct 5th roots of unity. Find the value of the product ∏k=15(2−zk)\prod_{k=1}^{5} (2-z_k).
    Answer: 31 The 5th roots of unity are the roots of the polynomial equation P(x)=x5−1=0P(x) = x^5 - 1 = 0. Since z1,z2,z3,z4,z5z_1, z_2, z_3, z_4, z_5 are the five roots of this polynomial, we can factorize P(x)P(x) as: x5−1=(x−z1)(x−z2)(x−z3)(x−z4)(x−z5)x^5 - 1 = (x-z_1)(x-z_2)(x-z_3)(x-z_4)(x-z_5). The product we want to find is (2−z1)(2−z2)(2−z3)(2−z4)(2−z5)(2-z_1)(2-z_2)(2-z_3)(2-z_4)(2-z_5), which is the value of the polynomial P(x)P(x) when x=2x=2. Substituting x=2x=2 into the equation x5−1x^5 - 1: P(2)=25−1=32−1=31P(2) = 2^5 - 1 = 32 - 1 = 31. Therefore, the value of the product is 31.
  14. Let z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta. By considering the binomial expansion of z5z^5, find an expression for cos⁡(5θ)\cos(5\theta).
    Answer: cos⁡5θ−10cos⁡3θsin⁡2θ+5cos⁡θsin⁡4θ\cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta By De Moivre's theorem, z5=(cos⁡θ+isin⁡θ)5=cos⁡(5θ)+isin⁡(5θ)z^5 = (\cos\theta + i\sin\theta)^5 = \cos(5\theta) + i\sin(5\theta). By the binomial theorem, (cos⁡θ+isin⁡θ)5=∑k=05(5k)(cos⁡θ)5−k(isin⁡θ)k(\cos\theta + i\sin\theta)^5 = \sum_{k=0}^{5} \binom{5}{k} (\cos\theta)^{5-k} (i\sin\theta)^k. Expanding this gives: (50)cos⁡5θ+(51)cos⁡4θ(isin⁡θ)+(52)cos⁡3θ(isin⁡θ)2+(53)cos⁡2θ(isin⁡θ)3+(54)cos⁡θ(isin⁡θ)4+(55)(isin⁡θ)5\binom{5}{0}\cos^5\theta + \binom{5}{1}\cos^4\theta(i\sin\theta) + \binom{5}{2}\cos^3\theta(i\sin\theta)^2 + \binom{5}{3}\cos^2\theta(i\sin\theta)^3 + \binom{5}{4}\cos\theta(i\sin\theta)^4 + \binom{5}{5}(i\sin\theta)^5. This becomes: 1⋅cos⁡5θ+5icos⁡4θsin⁡θ−10cos⁡3θsin⁡2θ−10icos⁡2θsin⁡3θ+5cos⁡θsin⁡4θ+isin⁡5θ1\cdot\cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta. cos⁡(5θ)\cos(5\theta) is the real part of this expansion. Grouping the real terms: Re(z5)=cos⁡5θ−10cos⁡3θsin⁡2θ+5cos⁡θsin⁡4θ\text{Re}(z^5) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta.
  15. Given that zz is a complex number such that z+1z=2cos⁡θz + \frac{1}{z} = 2\cos\theta, find an expression for zn+1znz^n + \frac{1}{z^n}, where nn is a positive integer.
    Answer: 2cos⁡(nθ)2\cos(n\theta) The given equation is z+1z=2cos⁡θz + \frac{1}{z} = 2\cos\theta. Multiplying by zz gives z2+1=2zcos⁡θz^2 + 1 = 2z\cos\theta, which can be rearranged into a quadratic equation: z2−(2cos⁡θ)z+1=0z^2 - (2\cos\theta)z + 1 = 0. Using the quadratic formula, z=2cos⁡θ±4cos⁡2θ−42=cos⁡θ±cos⁡2θ−1=cos⁡θ±−sin⁡2θ=cos⁡θ±isin⁡θz = \frac{2\cos\theta \pm \sqrt{4\cos^2\theta - 4}}{2} = \cos\theta \pm \sqrt{\cos^2\theta - 1} = \cos\theta \pm \sqrt{-\sin^2\theta} = \cos\theta \pm i\sin\theta. Let's choose z=cos⁡θ+isin⁡θ=eiθz = \cos\theta + i\sin\theta = e^{i\theta}. Then 1z=z−1=e−iθ=cos⁡θ−isin⁡θ\frac{1}{z} = z^{-1} = e^{-i\theta} = \cos\theta - i\sin\theta. Using De Moivre's theorem, zn=(eiθ)n=einθ=cos⁡(nθ)+isin⁡(nθ)z^n = (e^{i\theta})^n = e^{in\theta} = \cos(n\theta) + i\sin(n\theta). Similarly, 1zn=z−n=e−inθ=cos⁡(−nθ)+isin⁡(−nθ)=cos⁡(nθ)−isin⁡(nθ)\frac{1}{z^n} = z^{-n} = e^{-in\theta} = \cos(-n\theta) + i\sin(-n\theta) = \cos(n\theta) - i\sin(n\theta). Therefore, zn+1zn=(cos⁡(nθ)+isin⁡(nθ))+(cos⁡(nθ)−isin⁡(nθ))=2cos⁡(nθ)z^n + \frac{1}{z^n} = (\cos(n\theta) + i\sin(n\theta)) + (\cos(n\theta) - i\sin(n\theta)) = 2\cos(n\theta). Choosing z=cos⁡θ−isin⁡θz = \cos\theta - i\sin\theta would yield the same result.
  16. Let z=(sin⁡(π6)+icos⁡(π6))5z = (\sin(\frac{\pi}{6}) + i\cos(\frac{\pi}{6}))^5. Find the principal argument of zz.
    Answer: −π3-\frac{\pi}{3} When you encounter a complex number raised to a power, you're dealing with De Moivre's theorem, which states that for a complex number in polar form r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta), raising it to power nn gives rn(cos⁡(nθ)+isin⁡(nθ))r^n(\cos(n\theta) + i\sin(n\theta)). First, notice that the given complex number z=(sin⁡(π6)+icos⁡(π6))5z = (\sin(\frac{\pi}{6}) + i\cos(\frac{\pi}{6}))^5 isn't in standard polar form. The standard form is r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta), but we have sine first, then cosine. Using the identity sin⁡θ=cos⁡(π2−θ)\sin\theta = \cos(\frac{\pi}{2} - \theta) and cos⁡θ=sin⁡(π2−θ)\cos\theta = \sin(\frac{\pi}{2} - \theta), we can rewrite this as: sin⁡(π6)+icos⁡(π6)=cos⁡(π3)+isin⁡(π3)\sin(\frac{\pi}{6}) + i\cos(\frac{\pi}{6}) = \cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}) Since ∣sin⁡(π6)+icos⁡(π6)∣=1|\sin(\frac{\pi}{6}) + i\cos(\frac{\pi}{6})| = 1, we have z=(cos⁡(π3)+isin⁡(π3))5z = (\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))^5. By De Moivre's theorem: z=cos⁡(5π3)+isin⁡(5π3)z = \cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3}) The argument is 5π3\frac{5\pi}{3}. However, since 5π3>π\frac{5\pi}{3} > \pi, we need the principal argument (between −π-\pi and π\pi): 5π3−2π=−π3\frac{5\pi}{3} - 2\pi = -\frac{\pi}{3}. Answer D is correct: −π3-\frac{\pi}{3}. Answer A (−5π6-\frac{5\pi}{6}) incorrectly applies the original angle without conversion. Answer B (5π6\frac{5\pi}{6}) makes the same error but with wrong sign. Answer C (π3\frac{\pi}{3}) uses the original angle instead of the fifth power. Strategy tip: Always convert unusual trigonometric forms to standard cos⁡+isin⁡\cos + i\sin format before applying De Moivre's theorem, and remember to reduce arguments to the principal range.
  17. Consider the equation z4=−8−8i3z^4 = -8 - 8i\sqrt{3}. How many of its roots lie in the upper half of the complex plane (i.e., have a positive imaginary part)?
    Answer: 2 First, convert w=−8−8i3w = -8 - 8i\sqrt{3} to polar form. The modulus is ∣w∣=(−8)2+(−83)2=64+64⋅3=64(1+3)=256=16|w| = \sqrt{(-8)^2 + (-8\sqrt{3})^2} = \sqrt{64 + 64 \cdot 3} = \sqrt{64(1+3)} = \sqrt{256} = 16. The argument θ\theta satisfies cos⁡θ=−8/16=−1/2\cos\theta = -8/16 = -1/2 and sin⁡θ=−83/16=−3/2\sin\theta = -8\sqrt{3}/16 = -\sqrt{3}/2. This corresponds to an angle in the third quadrant, so θ=−2π/3\theta = -2\pi/3 (or 4π/34\pi/3). We need to solve z4=16ei(−2π/3+2kπ)z^4 = 16e^{i(-2\pi/3 + 2k\pi)}. The roots are zk=164ei(−2π/3+2kπ)/4=2ei(−π/6+kπ/2)z_k = \sqrt[4]{16} e^{i(-2\pi/3 + 2k\pi)/4} = 2e^{i(-\pi/6 + k\pi/2)} for k=0,1,2,3k=0, 1, 2, 3. Let's find the arguments of the four roots: For k=0k=0: arg⁡(z0)=−π/6\arg(z_0) = -\pi/6. This is in Q4, imaginary part is negative. For k=1k=1: arg⁡(z1)=−π/6+π/2=2π/6=π/3\arg(z_1) = -\pi/6 + \pi/2 = 2\pi/6 = \pi/3. This is in Q1, imaginary part is positive. For k=2k=2: arg⁡(z2)=−π/6+π=5π/6\arg(z_2) = -\pi/6 + \pi = 5\pi/6. This is in Q2, imaginary part is positive. For k=3k=3: arg⁡(z3)=−π/6+3π/2=8π/6=4π/3\arg(z_3) = -\pi/6 + 3\pi/2 = 8\pi/6 = 4\pi/3. The principal argument is −2π/3-2\pi/3. This is in Q3, imaginary part is negative. The arguments of the roots are between 0 and π\pi for k=1k=1 and k=2k=2. Therefore, exactly two roots have a positive imaginary part.
  18. The roots of the equation z5=−32z^5 = -32 form the vertices of a regular pentagon in the complex plane. Find the area of this pentagon.
    Answer: 10sin⁡(2π/5)10\sin(2\pi/5) The equation is z5=−32z^5 = -32. In polar form, −32=32eiπ-32 = 32e^{i\pi}. The roots are given by zk=325exp⁡(iπ+2kπ5)z_k = \sqrt[5]{32} \exp\left(i\frac{\pi + 2k\pi}{5}\right) for k=0,1,2,3,4k=0, 1, 2, 3, 4. The modulus of each root is r=325=2r = \sqrt[5]{32} = 2. This means the vertices of the pentagon lie on a circle of radius 2 centered at the origin. The area of a regular n-gon inscribed in a circle of radius rr is given by the formula A=12nr2sin⁡(2πn)A = \frac{1}{2}nr^2\sin(\frac{2\pi}{n}). In this case, n=5n=5 and r=2r=2. Substituting these values, we get: A=12(5)(22)sin⁡(2π5)=12(5)(4)sin⁡(2π5)=10sin⁡(2π5)A = \frac{1}{2}(5)(2^2)\sin(\frac{2\pi}{5}) = \frac{1}{2}(5)(4)\sin(\frac{2\pi}{5}) = 10\sin(\frac{2\pi}{5}).
  19. Let z=1−i3z = 1 - i\sqrt{3}. Find the value of z10z^{10}.
    Answer: −512+512i3-512 + 512i\sqrt{3} First, convert z=1−i3z = 1 - i\sqrt{3} to polar form r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta). The modulus is r=∣z∣=12+(−3)2=1+3=2r = |z| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2. The argument θ\theta satisfies cos⁡θ=1/2\cos\theta = 1/2 and sin⁡θ=−3/2\sin\theta = -\sqrt{3}/2. This is in the fourth quadrant, so θ=−π/3\theta = -\pi/3. Thus, z=2e−iπ/3z = 2e^{-i\pi/3}. By De Moivre's theorem, z10=(2e−iπ/3)10=210e−i10π/3z^{10} = (2e^{-i\pi/3})^{10} = 2^{10}e^{-i10\pi/3}. 210=10242^{10} = 1024. The argument is −10π/3-10\pi/3. To find the principal argument, we add multiples of 2π2\pi. −10π/3+2(2π)=−10π/3+12π/3=2π/3-10\pi/3 + 2(2\pi) = -10\pi/3 + 12\pi/3 = 2\pi/3. So, z10=1024(cos⁡(2π/3)+isin⁡(2π/3))z^{10} = 1024(\cos(2\pi/3) + i\sin(2\pi/3)). Converting back to Cartesian form: z10=1024(−1/2+i3/2)=−512+512i3z^{10} = 1024(-1/2 + i\sqrt{3}/2) = -512 + 512i\sqrt{3}.

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